Question

Let $h\left(x\right)=\left\{\begin{array}{l}xifx<0\\ x^{2}if0<x\le 2\\ 8-xifx>2\end{array}\right.$

Evaluate each limit if it exists.

$\begin{array}{l}(\text{i)}\underset{x\to 0^{+}}{\lim }h\left(x\right)\text{(iv)}\underset{x\to 2^{-}}{\lim }h\left(x\right)\\ \text{(ii)}\underset{x\to 0}{\lim }h\left(x\right)\text{(v)}\underset{x\to 2^{+}}{\lim }h\left(x\right)\\ \text{(iii)}\underset{x\to 1}{\lim }h\left(x\right)\text{(vi)}\underset{x\to 2}{\lim }h\left(x\right)\end{array}$

Sketch the graph of h .

Short answer

$\left(a\right)$

$\text{(i)}$0

$\text{(ii)}$0

$\text{(iii)}$1

$\text{(iv)}$4

$\text{(v)6}$

$\text{(vi)}$Limit doesn’t exist.

Step-by-step solution

Step 1. Given information.

Given function,

$h\left(x\right)=\left\{\begin{array}{l}xifx<0\\ x^{2}if0<x\le 2\\ 8-xifx>2\end{array}\right.$

Step 2. Find the limit.

$\text{(i)}$Given $\underset{x\to 0^{+}}{\lim }h\left(x\right)$

Plug the value $x=0$in$x^2$

We get, $0^2=0$

$\text{(ii)}$Given $\underset{x\to 0}{\lim }h\left(x\right)$

Here,

Left-hand limit

$\begin{array}{c}=x\\ =0\end{array}$

Right-hand limit

$\begin{array}{c}=x^2\\ =0^2\\ =0\end{array}$

We see that $L.H.L.=R.H.L.$

Therefore, limit exists and value of the limit $=0$.

$\text{(iii)}$Given,$\underset{x\to 1}{\lim }h\left(x\right)$

Plug the value $x=1$in $x^2$

We get, $1^2=1$

$\text{(iv)}$Given, $\underset{x\to 2^{-}}{\lim }h\left(x\right)$

Plug the value $x=2$in $x^2$

We get, $2^2=4$

$\text{(v)}$Given $\underset{x\to 2^{+}}{\lim }h\left(x\right)$

Plug the value $x=2$in$8-x$

We get, $8-2=6$

$\text{(vi)}$Given,$\underset{x\to 2}{\lim }h\left(x\right)$

Here,

Left-hand limit

$\begin{array}{c}=x^2\\ =2^2\\ =4\end{array}$

Right-hand limit

$\begin{array}{c}=8-x\\ =8-2\\ =6\end{array}$

We see that, $L.H.L.\ne R.H.L.$

Therefore, limit doesn’t exist.

$\left(b\right)$

Step 3. Sketch the graph.

Now graph of$h$ drawn below,